Answer:
Percent yield = 61.58%
Explanation:
C4H9OH + NaBr + H2SO4 --> C4H9Br + NaHSO4 + H2O
From the reaction;
1 mol of C4H9OH reacts with 1 mol of NaBr and 1 mol of H2SO4 to form 1 mol of C4H9Br
We have to obtain the limiting reagent. We do this by calculatig the mols of each reactant in the reaction;
Number of moles = Mass / Molar mass
C4H9OH; 15 / 74g/mol = 0.2027
NaBr; 22.4 / 102.89 = 0.2177
H2SO4; 32.7 / 98 = 0.3336
In this case, the limiting reagent is C4H9OH, it determines the amount of product formed.
From the stoichometry of the reaction; one mole of C4H9OH forms one mole of C4H9Br. This means 0.2027 mol of C4H9Br was formed.
Mass = Number of moles * Molar mass
Mass of C4H9Br formed = 0.2027 * 137
Mass = 27.77g
Percent yield = (Practical yield / theoretical yeield ) * 100%
Percent Yield = 17.1 / 27.77 * 100
Percent yield = 61.58%
